Multivariable calculus

Double Integral of sin^2 x cos^2 y Over a Square

Q: Does the answer change if the integrand is cos^2 x sin^2 y instead?

No. Each factor still integrates to \pi/2 over [0,\pi], so the product is again \pi^2/4. The symmetry comes from integrating over a full half period in both variables.

Q: Can I use sin^2 + cos^2 = 1 to simplify this integrand?

Not here. That identity applies to \sin^{2}\theta + \cos^{2}\theta for the same angle. In \sin^{2}x\,\cos^{2}y the angles differ and the terms are multiplied, so the identity does not touch it.

The double integral of sin(x) squared times cos(y) squared over the square with x and y each from 0 to pi equals pi squared over 4, about 2.4674. The integrand separates, and each single integral of a squared trig function over 0 to pi equals pi/2.

f(x,y) = \sin^{2}x\,\cos^{2}y

\dfrac{\pi^{2}}{4}

Numerically 2.467401, confirmed by quadrature on every build.

Separate, then kill each square with a half-angle identity

The integrand is a product of a function of $x$ alone and a function of $y$ alone, and the region is a rectangle, so the double integral splits into two independent single integrals.

\iint_{[0,\pi]^{2}} \sin^{2}x\,\cos^{2}y\,dA = \left(\int_{0}^{\pi}\sin^{2}x\,dx\right)\left(\int_{0}^{\pi}\cos^{2}y\,dy\right)

Neither square has an obvious antiderivative, so lower the power with the half-angle identities $\sin^{2}x = \frac{1-\cos 2x}{2}$ and $\cos^{2}y = \frac{1+\cos 2y}{2}$ .

\int_{0}^{\pi}\sin^{2}x\,dx = \int_{0}^{\pi}\frac{1-\cos 2x}{2}\,dx = \left[\frac{x}{2} - \frac{\sin 2x}{4}\right]_{0}^{\pi} = \frac{\pi}{2}

The cosine version works the same way and also gives $\pi/2$ , since $\sin 2y$ vanishes at both ends. Multiplying gives $\frac{\pi}{2}\cdot\frac{\pi}{2} = \frac{\pi^{2}}{4} \approx 2.4674$ .

The mistake: applying the power rule to sin^2 x

The most common wrong answer here comes from treating $\sin^{2}x$ as if it were $u^{2}$ with $du = dx$ , giving an antiderivative of $\frac{\sin^{3}x}{3}$ .

\frac{d}{dx}\left(\frac{\sin^{3}x}{3}\right) = \sin^{2}x\cos x \neq \sin^{2}x

The stray $\cos x$ is the chain-rule factor that is missing from the integrand. Worse, that fake antiderivative evaluates to $0$ at both $x = 0$ and $x = \pi$ , so it reports the whole integral as $0$ rather than $\pi/2$ .

A zero answer should have set off alarms on its own: $\sin^{2}x$ is non-negative and not identically zero, so its integral over an interval of positive length must be strictly positive.

Why both squares average to one half

There is a faster route worth knowing. On $[0,\pi]$ both $\sin^{2}$ and $\cos^{2}$ have average value $\frac{1}{2}$ : they add to $1$ at every point, and each is the other shifted by $\pi/2$ , which changes nothing across a full period of length $\pi$ . Equal averages summing to $1$ must each be $\frac{1}{2}$ .

\frac{1}{\pi}\int_{0}^{\pi}\sin^{2}x\,dx = \frac{1}{2} \quad \Longrightarrow \quad \int_{0}^{\pi}\sin^{2}x\,dx = \frac{\pi}{2}

So the double integral is $\frac{1}{2}\cdot\frac{1}{2}$ times the base area $\pi^{2}$ , which is $\frac{\pi^{2}}{4}$ . The averaging argument does need the right interval. Since $\int_{a}^{b}\sin^{2}x\,dx = \frac{b-a}{2} - \left[\frac{\sin 2x}{4}\right]_{a}^{b}$ , the average is exactly $\frac{1}{2}$ only when $\sin 2a = \sin 2b$ . Every interval of length $\pi$ qualifies, and so does $[0,\pi/2]$ , but over $[0,1]$ the leftover $\frac{\sin 2}{4} \approx 0.227$ pulls the average down to $0.273$ .

Frequently asked questions

Does the answer change if the integrand is cos^2 x sin^2 y instead?

No. Each factor still integrates to $\pi/2$ over $[0,\pi]$ , so the product is again $\pi^2/4$ . The symmetry comes from integrating over a full half period in both variables.

Can I use sin^2 + cos^2 = 1 to simplify this integrand?

Not here. That identity applies to $\sin^{2}\theta + \cos^{2}\theta$ for the same angle. In $\sin^{2}x\,\cos^{2}y$ the angles differ and the terms are multiplied, so the identity does not touch it.